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Chapter 10: Problem 115

For the reaction \(\mathrm{A} \longrightarrow\) Products, it is found that the rate of reaction increases by a factor of \(6.25\), when the concentration of \(\mathrm{A}\) is increased by a factor of \(2.5 .\) The order of reaction with respect to \(\mathrm{A}\) is (a) \(0.5\) (b) 1 (c) 2 (d) 3

Short Answer

Expert verified
The order of reaction with respect to A is 2.

Step by step solution

01

Write the Rate Law

For a reaction of the form \(\mathrm{A} \rightarrow\) Products, the rate law can be expressed as \(\text{Rate} = k [\mathrm{A}]^n\), where \([\mathrm{A}]\) is the concentration of A, \(k\) is the rate constant, and \(n\) is the order of the reaction with respect to A.
02

Understand the Given Change in Rate

It is given that when the concentration of A increases by a factor of \(2.5\), the rate of reaction increases by a factor of \(6.25\). This implies that the new rate can be written as \(6.25 \times \text{Rate} = k (2.5 \times [\mathrm{A}])^n\).
03

Set Up the Equation

Using the rate law and given information, we can set up the equation: \(6.25 = (2.5)^n\).
04

Solve for the Order \(n\)

To solve for \(n\), we take the logarithm of both sides: \[\log(6.25) = n \cdot \log(2.5)\]. Calculate \(\log(6.25)\) and \(\log(2.5)\) to find \(n\).
05

Calculation

Calculate the values: \(\log(6.25) = 0.796\) and \(\log(2.5) = 0.398\). Then divide to find \(n\): \(\frac{0.796}{0.398} = 2\).
06

Conclusion on the Reaction Order

The calculated order \(n\) is \(2\), therefore, the order of reaction with respect to A is \(2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law
The rate law is a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. For a reaction in the form
  • \( ext{A} \rightarrow \text{Products} \)
the rate law can be expressed as:
  • \(\text{Rate} = k [\text{A}]^n\)
In this equation,
  • \(\text{Rate}\) is the speed of the reaction.
  • \(k\) is the rate constant, a fixed value at a given temperature
  • \([\text{A}]\) represents the concentration of reactant A
  • \(n\) is the reaction order with respect to A.
The order of the reaction with respect to a reactant indicates how the reaction rate is affected by that reactant's concentration. If \(n = 0\), the rate is independent of \[A\]. If \(n\) equals 1, the rate is directly proportional, and if \(n = 2\), the rate varies with the square of the concentration of A.
Understanding and determining the correct rate law is crucial for predicting how a reaction will proceed under different conditions.
Chemical Kinetics
Chemical kinetics is the study of the rates of chemical reactions, revealing how different variables, such as concentration, temperature, and catalysts, influence reaction speed. It delves into the mechanism of the reaction鈥攖he step-by-step sequence of elementary reactions.
Several key terms are central to chemical kinetics:
  • **Reaction rate:** Speed at which reactant concentrations decrease or product concentrations increase.
  • **Rate constant (k):** A proportionality constant in the rate law equation, specific to a given reaction at a given temperature.
Chemical kinetics not only helps us understand current reaction conditions but also allows us to control reactions. For instance, in industrial applications, optimizing these conditions can maximize yield and efficiency of chemical processes.
By studying kinetics, chemists can make predictions about how a reaction's yield will change when modifying the concentration of reactants or catalysts. This essential insight is foundational for research and development in chemistry.
Concentration Effect on Reaction Rate
The concentration of reactants is a crucial factor affecting reaction rates. As a general rule, increasing the concentration of reactants tends to increase the rate of reaction. This is because more reactant molecules are present, leading to a higher probability of collisions, and thus, more frequent successful interactions required for the reaction.
  • **Direct Proportionality:** In first-order reactions, the rate is directly proportional to the concentration of one reactant. For example, if the concentration doubles, the rate doubles.
  • **Exponential Relationship:** In second-order reactions, if the concentration is doubled, the rate increases by a factor of four \((2^2)\).
The degree of dependence varies based on the order assigned to a particular reactant in the rate law.
By experimentally determining the rate law, one can predict how changes in concentration affect the rate, which allows chemists to manipulate conditions to achieve desired outcomes. This understanding is key in both academic and industrial applications of chemical reactions.

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Most popular questions from this chapter

Which of the following statements are correct about half-life period? (1) time required for \(99.9 \%\) completion of a reaction is 100 times the half-life period (2) time required for \(75 \%\) completion of a first-order reaction is double the half-life of the reaction (3) average life \(=1.44\) times the half-life for firstorder reaction

From the following data for the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) $$ \begin{array}{cccc} \hline[\mathrm{A}] & {[\mathrm{B}]} & \text { Initial rate } & \left(\mathrm{mol} \mathrm{L}^{-1} \mathbf{s}^{-1}\right) \\ \mathrm{molL}^{-1} & \mathrm{molL}^{-1} & \mathbf{3 0 0} \mathrm{K} & \mathbf{3 2 0} \mathrm{K} \\ \hline 2.5 \times 10^{-4} & 3.0 \times 10^{-5} & 5.0 \times 10^{-4} & 2.0 \times 10^{-3} \\ 5.0 \times 10^{-4} & 6.0 \times 10^{-5} & 4.0 \times 10^{-3} & \- \\ 1.0 \times 10^{-3} & 6.0 \times 10^{-5} & 1.6 \times 10^{-2} & \- \\ \hline \end{array} $$ Calculate the rate equation (a) \(\mathbf{r}=k[\mathrm{~B}]^{1}\) (b) \(\mathrm{r}=k[\mathrm{~A}]^{2}\) (c) \(\mathbf{r}=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{1}\) (d) \(\mathrm{r}=\mathrm{A}[\mathrm{A}][\mathrm{B}]\)

The basic theory of Arrhenius equation is that 12 (1) activation energy and pre-exponential factors are always temperature independent (2) the number of effective collisions is proportional to the number of molecule above a certain thresh old energy. (3) as the temperature increases, the number of molecules with energies exceeding the threshold energy increases. (4) the rate constant in a function of temperature (a) 2,3 and 4 (b) 1,2 and 3 (c) 2 and 3 (d) 1 and 3

The reaction \(\mathrm{X} \longrightarrow\) Product follows first-order kinetics, in 40 minutes, the concentration of \(X\) changes from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\), then the rate of reaction when concentration of \(X\) is \(0.01 \mathrm{M}\) is? (a) \(3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (b) \(1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}\) (c) \(1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (d) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\)

The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at \(25^{\circ} \mathrm{C}\) are \(3.0 \times 10^{-4} \mathrm{~s}^{-1}, 104.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(6 \times 10^{14} \mathrm{~s}^{-1}\) respectively. The value of the rate constant as \(\mathrm{T} \longrightarrow \infty\) is (a) \(2.0 \times 10^{18} \mathrm{~s}^{-1}\) (b) \(6.0 \times 10^{14} \mathrm{~s}^{-1}\) (c) infinity (d) \(3.6 \times 10^{30} \mathrm{~s}^{-1}\)
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